Answer:
28 m/s^2
Explanation:
distance, s = 14 m
time, t = 2 - 1 = 1 s
initial velocity, u = 0 m/s
Let a be the acceleration.
Use third equation of motion


a = 28 m/s^2
Thus, the acceleration is 28 m/s^2.
The electrostatic force between two charges q1 and q2 is given by

where

is the Coulomb's constant and r is the distance between the two charges.
If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
Answer: 8.6 µm
Explanation:
At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:
Ey =Emax cos (kx-ωt)
So, we can write the following equality:
ω= 2.2 1014 rad/sec
The angular frequency and the linear frequency are related as follows:
f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec
In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.
The wavelength, speed and frequency, are related by this equation:
λ = c/f
λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.
Answer:
Charge of particle 2, 
Explanation:
Given that,
Charge 1, 
The distance between charges, r = 0.241 m
Force experienced by particle 1, F = 3.44 N
We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :




or

So, the magnitude of electric charge 2 is
. Since, the force is attractive then the magnitude of charge 2 must be negative.